4. Practice with Pauli matrix operators: Consider the three 2 x 2 Pauli matrices of, oy, of. These matrices have the following properties: they are unitary, Hermitian, square to identity, and satisfy o"oy = 10%, 636% = io", 0 0 = ios. Using these properties, show the following: (a) eldon = cos(0) 1 + isin(0) on where n is a three-component unit-vector, and O is a real number. (b) (8 . a) (6 . b) = a . b1 + io . (a x b), where a, b are three-component vectors. 2 (c) tr (8 . a) (8 . b) (8 . c) = 2ia . (bx c), where a, b, care three-component vectors, and 'tr' denotes trace. (d) Lizx,y,2 08075 = 20a,868,y - da,Boy,8 where o'. denotes the (a, B) element of matrix o', e.g., of.1 = 1, 0 2 =0 etc. Hint: Consider a 2 x 2 matrix A and express it in terms of Pauli matrices as A = c1 +a . o, where c is a constant, and a is a vector, both of which you can determine in terms of matrix A using Pauli matrices' aforementioned properties. Since this equality holds for any 2 x 2 matrix A, it leads to the identity that you are asked to prove.The Pauli matrices are defined as: Given a 2x2 matrix A, we can express it in terms of Pauli matrices as: A= cl + a .a where I is the identity matrix, c is a constant, and a is a vector. We can determine c and a in terms of matrix A: c = ITr(A) a = *Tr(Ad) Substituting these into the expression for A, we get:A = }Tr(A)I + =Tr(Ao) . a Now, we can substitute this into the given equation and simplify: oi (-Tr(A)I + :Tr(Ao) . )as(Tr(A)I + -Tr(Ao) . d),s = 260,86BY - ba,poy,6 Simplifying, we get: "Tr(A)' IaBlya + 'Tr(A) dag . da + Tr(Ao) . dapTr(Ao) . dys = 26a,868, This is the identity we were asked to prove